Problem: Simplify the following expression and state the condition under which the simplification is valid. $k = \dfrac{4r^2 - 40r + 36}{r^3 + 3r^2 - 4r}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ k = \dfrac {4(r^2 - 10r + 9)} {r(r^2 + 3r - 4)} $ $ k = \dfrac{4}{r} \cdot \dfrac{r^2 - 10r + 9}{r^2 + 3r - 4} $ Next factor the numerator and denominator. $ k = \dfrac{4}{r} \cdot \dfrac{(r - 1)(r - 9)}{(r - 1)(r + 4)}$ Assuming $r \neq 1$ , we can cancel the $r - 1$ $ k = \dfrac{4}{r} \cdot \dfrac{r - 9}{r + 4}$ Therefore: $ k = \dfrac{ 4(r - 9)}{ r(r + 4)}$, $r \neq 1$